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Ch 30: Inductance
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 30: InductanceProblem 11b
Chapter 30, Problem 11b

Inductance of a Solenoid. A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

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Understand the formula for the inductance of a solenoid: \( L = \frac{\mu_0 N^2 A}{l} \), where \( L \) is the inductance, \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( l \) is the length of the solenoid.
Calculate the cross-sectional area \( A \) of the solenoid using the formula for the area of a circle: \( A = \pi r^2 \). The radius \( r \) is half of the diameter, so \( r = \frac{0.150 \text{ cm}}{2} \). Convert the radius to meters before calculating the area.
Convert the length of the solenoid from centimeters to meters. The length \( l \) is given as 5.00 cm, which is equivalent to 0.050 meters.
Substitute the values into the inductance formula. Use \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \) for the permeability of free space, \( N = 50 \) for the number of coils, and the calculated values for \( A \) and \( l \).
Simplify the expression to find the self-inductance \( L \) of the solenoid. Ensure all units are consistent, and perform the necessary arithmetic operations to complete the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductance

Inductance is a property of an electrical conductor that quantifies its ability to induce an electromotive force (EMF) when the current flowing through it changes. It is measured in henries (H) and is crucial for understanding how solenoids and other inductive components behave in circuits, affecting energy storage and transfer.
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Solenoid

A solenoid is a coil of wire designed to create a magnetic field when an electric current passes through it. The inductance of a solenoid depends on its physical characteristics, such as the number of turns, length, and cross-sectional area, as well as the permeability of the core material. Solenoids are used in various applications, including electromagnets and inductors.
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Self-Inductance Formula for Solenoids

The self-inductance of a solenoid can be calculated using the formula L = (μ₀ * N² * A) / l, where L is the inductance, μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid. This formula helps determine how the solenoid will behave in an electrical circuit, particularly in terms of energy storage and magnetic field generation.
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Related Practice
Textbook Question

The inductor shown in Fig. E30.11 has inductance 0.260 H and carries a current in the direction shown. The current is changing at a constant rate. The potential between points a and b is Vab = 1.04 V, with point a at higher potential. Is the current increasing or decreasing?

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Textbook Question

An air-filled toroidal solenoid has a mean radius of 15.0 cm and a cross-sectional area of 5.00 cm2. When the current is 12.0 A, the energy stored is 0.390 J. How many turns does the winding have?

Textbook Question

At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

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Textbook Question

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

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Textbook Question

A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is 3.25 × 10-3 Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?

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Textbook Question

A solenoid 25.0 cm long and with a cross-sectional area of 0.500 cm2 contains 400 turns of wire and carries a current of 80.0 A. Calculate: the total energy contained in the coil's magnetic field (assume the field is uniform);

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